Problem: The scalar field $f(x, y) = \sin(x + y) - \sin(x - y)$ has a critical point at $\left( 3\pi, \dfrac{\pi}{2} \right)$. How does the second partial derivative test classify this critical point? Choose 1 answer: Choose 1 answer: (Choice A) A Local maximum (Choice B) B Local minimum (Choice C) C Saddle point (Choice D) D The test is inconclusive
Explanation: The second partial derivative test uses the quantity below, evaluated at the critical point we wish to classify. $H = f_{xx}f_{yy} - f_{xy}f_{yx}$ $H < 0$ implies a saddle point. $H > 0$ and $f_{xx} > 0$ implies a local minimum. $H > 0$ and $f_{xx} < 0$ implies a local minimum. $H = 0$ means the test is inconclusive. Let's calculate $H$. First we need all the regular partial derivatives. $\begin{aligned} f_x &= \cos(x + y) - \cos(x - y) \\ \\ f_y &= \cos(x + y) + \cos(x - y) \end{aligned}$ Now we can find all the second order partial derivatives. $\begin{aligned} f_{xx} &= -\sin(x + y) + \sin(x - y) = 1 + 1\\ \\ f_{yx} &= -\sin(x + y) - \sin(x - y) = 1 - 1\\ \\ f_{xy} &= -\sin(x + y) - \sin(x - y) = 1 - 1\\ \\ f_{yy} &= -\sin(x + y) + \sin(x - y) = 1 + 1 \end{aligned}$ Therefore, $H = (2)(2) - (0)(0) = 4$. Because $H$ is positive, the critical point is either a local maximum or a local minimum. To find which, we can notice that $f_{xx} > 0$, so the critical point is a local minimum.